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Relativistic Correction

  1. May 7, 2014 #1
    For hydrogen atoms, all book take correction up to 1/c^2 where the perturbation is -P^4/8*m^3*c^2. And they go solving it by sandwiching p^4 term where they consider p^2 = 2m*(1+e^2/r). and they square it to solve for p^4.

    To get a better view of perturbation to first order please see attachment.


    What if I want it to the second order correction, that is to p^6? The additional perturbative term would be p^6/16*m^5*c^4. What should be done in this case?
     

    Attached Files:

  2. jcsd
  3. May 7, 2014 #2
    I think you can use the same technique. The only new term you need to evaluate is [itex]\langle n,\ell|\frac{1}{r^3}|n,\ell\rangle[/itex].
     
  4. May 7, 2014 #3
    And what to do with P^6?
     
  5. May 7, 2014 #4
    It's the same technique, you can write:
    $$
    \frac{p^6}{16m^5c^4}=\frac{1}{2m^2c^4}\left(\frac{p^2}{2m}\right)^3= \frac{1}{2m^2c^4}\left(\frac{p^2}{2m}-\frac{e^2}{r}+\frac{e^2}{r}\right)^3.
    $$
    You now realize that [itex]\frac{p^2}{2m}-\frac{e^2}{r}=H_0[/itex] and so:
    $$
    \frac{p^6}{16m^5c^4}=\frac{1}{2m^2c^4}\left(H_0^3+\frac{e^6}{r^3}+3H_0^2\frac{e^2}{r}+3H_0\frac{e^4}{r^2}\right),
    $$
    and the you exactly what is the eigenvalue for [itex]H_0[/itex].
     
  6. May 7, 2014 #5
    Ohhhh thank you very much!
     
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